WhaT are 2 consecutive odd numbers that have the product of 12,099?

2 Answers
Jun 27, 2015

Answer:

I found:
#109xx111=12099#
or
#-111xx-109=12099#

Explanation:

Consider the condition for your two consecutive odd numbers:
#(2n+1)xx(2n+3)=12,099#
rearranging:
#4n^2+8n-12096=0#
Using the Quadratic Formla you should get:
#n_1=-56#
#n_2=54#
So you have in #(2n+1)xx(2n+3)=12,099#:
#109xx111=12099#
or
#-111xx-109=12099#

Jun 27, 2015

Answer:

I got the same answer as Gio (of course), but set it up a little differently.

Explanation:

When asked to find odd (or even) numbers, it is often a good idea to use #2n+1# for an odd number (and #2n# for an even).

In a problem like this, I like to simply get an equation, find all solution and then pick out the odd (or even) solutions if there are any.

I notice that #12,099# is very close to #12,100#, And also recalling that #121 = 11^2#, so #121 xx 100 = 11^2 xx 10^2#

Of course consecutive odd numbers are exactly #2# away from each other, so

let's call our numbers #x-1# and #x+1#

We want #(x-1)(x+1) = 12,099#

So solve #x^2-1=12,099# (Which we've sort of already done!)

#x^2 = 12,100#

#x= +- 110#

So the numbers are:

#110-1 = 109# and #110+1 = 111# (yes, they are odd)

and

#-110-1 = -111# and #-110+1 = -109# (yes, they're odd too)